Prove that every n s.t. gcd(n,10) = 1 has a multiple which is made up of only 9's

Prove that every n s.t. gcd(n,10) = 1 has a multiple which is made up of only 9's
Proof:

Let us pick one 'n'.
When we divide 10^1, 10^2, 10^3 .. with 'n' there can be at max 'n' different remainders.
So there must exist integers a,b with a>b s.t. 10^a and 10^b have same remainders when divided by 'n'.
So
10^a - 10^b is divisible by 'n'.
10^b(10^(a-b) - 1) is div by n.
gcd(n,10) = 1 => 10^(a-b) - 1 is div by 'n' and this is entirely made up of 9s.
H.P.

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